A block of mass m is placed on a smooth block of mass `M = m` with the help of a spring as shown in the figure. A velocity `v_(0)` is given to upper block when spring is in natural length. Find maximum compression in the spring.

Options

A. `sqrt((m)/(2K)) V_(0)`
B. `sqrt((m)/(K)) v_(0)`
C. `sqrt((2m)/(K)) v_(0)`
D. `2 sqrt((m)/(K)) v_(0)`

Answer

A

Explanation

Compression will be maximum when relative velocity between blocks is zero.
Means both have Sam velocity apply conservation of momentum
`mv_(0) = (M + m) v`
`rArr v = (m)/(M + m) v_(0)`
Apply conservation of kinetic energy
`(1)/(2) mv^(2) + (1)/(2) Mv^(2) + (1)/(2) kx^(2) = (1)/(2) mv_(0)^(2)`
`rArr (m^(3)v_(0)^(2))/((M + m)^(2)) + (m^(2)Mv_(0)^(2))/((M + m)^(2)) + kx^(2) = mv_(0)^(2)`
`rArr kx^(2) = (m^(3) + m^(2)M)/((M + m)^(2)) v_(0)^(2) = mv_(0)^(2)`
`rArr kx^(2) = v_(0)^(2) [(m)/(p) - (m^(2) (M + m))/((M + m)^(2))]`
`= v_(0)^(2) [(M_(m) + m^(2) - m^(2))/(M + m)] = (M_(m)v_(0)^(2))/(M + m)`
`rArr x = sqrt((Mm)/(k(M + m))) v_(0)`